Custom decimal round

[=?Utf-8?B?THVpZ2k=?=]

Hi all,
I need to round a decimal value with a particular rule.
For example:

decimal a = 1.49m -1m
decimal a = 1.5m -1m
decimal a = 1.51m -2m

I've tried Math.Round but with the value 1.5 it returns me 2, which it is
not correct for my business rule.

How can I solve this problem?

Thanks in advance.
--
Luigi

[Jon Skeet [C# MVP]]

On Jul 1, 3:23*pm, Luigi <ciupazNoSpamGr [e-mail] rote:

I need to round a decimal value with a particular rule.
For example:

decimal a = 1.49m -1m
decimal a = 1.5m -1m
decimal a = 1.51m -2m

I've tried Math.Round but with the value 1.5 it returns me 2, which it is
not correct for my business rule.

How can I solve this problem?

I'd write a custom rounding method. It's reasonable simple if you have
a very specific requirement. In this case, something like:

public decimal Round (decimal value)
{
decimal floor = Math.Floor(valu e);
decimal ceiling = Math.Floor(valu e);
decimal midpoint = (floor+ceiling)/2;
return value <= midpoint ? floor : ceiling;
}

It's possible that there will be some weird problems around the very
largest numbers that decimals can store, but if your application
doesn't use those (and it's unlikely to) then you should be okay.

This is completely untested though - I strongly recommend writing unit
tests for it!

Jon

[Marc Gravell]

How about something cheeky like:

decimal a = Math.Ceiling(x-0.5M);

Obviously you'd need to think about -ves etc...

Marc

[SLL]

On Jul 1, 10:23*am, Luigi <ciupazNoSpamGr [e-mail] rote:

Hi all,
I need to round a decimal value with a particular rule.
For example:

decimal a = 1.49m -1m
decimal a = 1.5m -1m
decimal a = 1.51m -2m

I've tried Math.Round but with the value 1.5 it returns me 2, which it is
not correct for my business rule.

How can I solve this problem?

Thanks in advance.

--
Luigi

Try multiplying by 10*(number of decimals needed), used math.floor or
math.ceiling, then divide by same number 10*(number of decimals
needed)

[Marc Gravell]

Assuming you mean 10 raised-to-the-power-of (number of decimals), the OP
appears to want zero decimals. Taking the cited example 1.51, this would
give 1M, not 2M as desired.

Marc

[rossum]

On Tue, 1 Jul 2008 07:55:38 -0700 (PDT), "Jon Skeet [C# MVP]"
<sk***@pobox.co mwrote:

>On Jul 1, 3:23*pm, Luigi <ciupazNoSpamGr [e-mail] rote:

>I need to round a decimal value with a particular rule.
For example:

decimal a = 1.49m -1m
decimal a = 1.5m -1m
decimal a = 1.51m -2m

I've tried Math.Round but with the value 1.5 it returns me 2, which it is
not correct for my business rule.

How can I solve this problem?

I'd write a custom rounding method. It's reasonable simple if you have
a very specific requirement. In this case, something like:

public decimal Round (decimal value)
{
decimal floor = Math.Floor(valu e);
decimal ceiling = Math.Floor(valu e);

Did you mean "decimal ceiling = Math.Ceiling(va lue);"? rossum

decimal midpoint = (floor+ceiling)/2;
return value <= midpoint ? floor : ceiling;
}

It's possible that there will be some weird problems around the very
largest numbers that decimals can store, but if your application
doesn't use those (and it's unlikely to) then you should be okay.

This is completely untested though - I strongly recommend writing unit
tests for it!

Jon

[Jon Skeet [C# MVP]]

rossum <ro******@coldm ail.comwrote:

<snip>

decimal ceiling = Math.Floor(valu e);

Did you mean "decimal ceiling = Math.Ceiling(va lue);"? rossum

Yup. That's what I get for not testing it...

--
Jon Skeet - <sk***@pobox.co m>
Web site: http://www.pobox.com/~skeet
Blog: http://www.msmvps.com/jon_skeet
C# in Depth: http://csharpindepth.com

[=?Utf-8?B?THVpZ2k=?=]

Thank you all very much.
I'll try this one method:

decimal x = 1.51m;

decimal a = Math.Ceiling(x - 0.5M);

Luigi

[Marc Gravell]

Note that this is called the banker rounding

I'm fairly certain that the original question as posed doesn't relate
to banker's rounding; it is just .5 always rounding down.

Marc

[Michel Walsh]

Yes, sorry, I was not clear. The OP want a rounding down at 0.5, but GOT a
banker rounding, with Math.Round(x, 0).

Vanderghast, Access MVP
"Marc Gravell" <ma**********@g mail.comwrote in message
news:c1******** *************** ***********@56g 2000hsm.googleg roups.com...

>Note that this is called the banker rounding

I'm fairly certain that the original question as posed doesn't relate
to banker's rounding; it is just .5 always rounding down.

Marc